Estimation on a summation Suppose we have two vectors $x$ and $y$ in $\mathbb{R}^n$ that satify


*

*$\|x\|=\|y\|=1$

*$<x,y>=0$

*$\sum_{i=1}^{n}{x_{i}}=\sum_{i=1}^{n}{y_i}=0$
That is $x$ and $y$ are of norm 1, $x\perp y$ and $x,y \perp e$ where $e=(1,…,1)$. Here both the norm and the inner product are the familiar Euclidean one. My question is, in such case, do we have $$\sum_{1\leq i<j\leq n}{|x_{i}-x_{j}||y_{i}-y_{j}|}\leq n-1$$
When $n=3$ this inequality can be verified by pure calculation. I suppose this is true for all integers $n\geq 3$. Could anyone help me prove this or give a counter example? Thanks in advance!
 A: Here is a partial  answer.
You have, using the Cauchy-Schwartz Inequality
$$
2 \sum_{1\leq i<j\leq n}{|x_{i}-x_{j}||y_{i}-y_{j}|} \\
= \sum_{1\leq i\leq n}\sum_{1\leq j\leq n}{|x_{i}-x_{j}||y_{i}-y_{j}|} \\
\le (\sum_{1\leq i\leq n}\sum_{1\leq j\leq n}{|x_{i}-x_{j}|^2})^{1/2}(\sum_{1\leq i\leq n}\sum_{1\leq j\leq n}{|y_{i}-y_{j}|^2})^{1/2} \\
= (\sum_{1\leq i\leq n}\sum_{1\leq j\leq n}{(x_{i}^2-2 x_{i} x_{j} + x_{j}^2)})^{1/2}(\sum_{1\leq i\leq n}\sum_{1\leq j\leq n}{(y_{i}^2-2 y_{i} y_{j} + y_{j}^2)})^{1/2}\\
= (2n - 2 \sum_{1\leq i\leq n}\sum_{1\leq j\leq n}{x_{i} x_{j} })^{1/2}(2n - 2 \sum_{1\leq i\leq n}\sum_{1\leq j\leq n}{y_{i} y_{j} })^{1/2}\\
= (2n - 2 (\sum_{1\leq i\leq n}{x_{i}  })^2)^{1/2}
(2n - 2 (\sum_{1\leq i\leq n}{y_{i}  })^2)^{1/2} \\
= 2n
$$
So this is close, but $2n -2$ was required. Note that there is only one inequality in the chain of reasoning (Cauchy-Schwartz). It is  known that Cauchy-Schwartz holds with equality if and only if the two vectors $|x_{i}-x_{j}|$ and $|y_{i}-y_{j}|$ are in the same direction, i.e. if for all $i,j$:  $|x_{i}-x_{j}|= c |y_{i}-y_{j}|$ with some constant $c$. If this were the case, you had a contradiction to your inequality. 
I guess it is pretty unlikely that this happens since there is still the orthogonality condition which was never used in the above reasoning.
A: Following the proof @Andreas using CBS we have the following refinement :

Particular case :
If $p_i=1$ ,$\sum_{i\in J }a_jp_j=0$ ,$\sum_{i\in H }p_ia_i^2=2n$,$\sum_{j\in J}p_jb_j=2$, $P_j=2$ ,$\sum_{i\in J} p_ia_i=0$
Then we can take the value :
$$A=4n$$
Second example :
If $p_i=\frac{1}{n}$ ,$\sum_{i\in J }a_jp_j=0$ ,$\sum_{i\in H }p_ia_i^2=2n$,$\sum_{j\in J}p_jb_j=\frac{1}{n}$, $P_j=\frac{1}{n}$ ,$\sum_{i\in J} p_ia_i=0$
We can take the value :
$$A=2n$$
This result is almost equivalent to the proof @Andreas.
Reference :
@article{Dragomir2003ASO,
title={A Survey on Cauchy-Buniakowsky-Schwartz Type Discrete Inequalities},
author={Sever Silvestru Dragomir},
journal={Mathematics eJournal},
year={2003}
}
https://www.semanticscholar.org/paper/A-Survey-on-Cauchy-Buniakowsky-Schwartz-Type-Dragomir/4ef9912775e7c66ee1cb13b93270a8c8c79a67f4
